Common fixed point theorem of six self-mappings in Menger spaces using (CLR_ST) property

Definition 1.1

A real valued functionfon the set of real numbers is called a distribution function if it is non-decreasing, left continuous withinfu∈Rf(u)=0 and supu∈Rf(u)=1.

Definition 1.2

([6]). LetXbe a non-empty set and letLdenote the set of all distribution functions defined onX, i.e., L = {F_{x, y} : x, y ∈ X}. An ordered pair (X, F) is called a probabilistic metric space (for short, PM-space) whereFis a mapping fromX × XintoLif, for every pair (x, y) ∈ X, a distribution functionF_{x, y}is assumed to satisfy the following four conditions:

1. F_{x, y}(u) = 1 for allu > 0, if and only ifx = y;

2. F_{x, y}(u) = F_{y, x}(u);

3. F_{x, y}(0) = 0;

4. IfF_{x, y}(u₁) = 1 andF_{y, z}(u₂) = 1, thenF_{x, z}(u₁ + u₂) = 1 for allx, y, zinXandu₁, u₂ ≥ 0.

Definition 1.3

([14]). At-norm is a function t : [0, 1] × [0, 1] → [0, 1] which satisfies the following conditions:

1. t(a, 1) = a, t(0, 0) = 0;

2. t(a, b) = t(b, a);

3. t(c, d) ≥ t(a, b) forc ≥ a, d ≥ b;

4. t(t(a, b), c) = t(a, t(b, c)) for alla, b, cin [0, 1].

Definition 1.4

([14]). A Menger probabilistic metric space (X, F, t)(for short, Menger-space) is an ordered triple, wheretis at-norm, and (X, F) is a probabilistic metric space which satisfies the following condition:

Definition 1.5

1. The pair (A, S) of self mappings of a Menger space (X, F, t) is said to satisfy the property (E.A) [15] if there exists a sequence {x_n} inXsuch that

   limn→∞Axn=limn→∞Sxn=z,  for some  z∈X.

2. Two pairs (A, S) and (B, T) of self mappings of a Menger space (X, F, t) are said to satisfy the property (E.A) [15] if there exists two sequences {x_n}, {y_n} inXsuch that

   limn→∞Axn=limn→∞Sxn=limn→∞Byn=limn→∞Tyn=z,  for some  z∈X.

3. The pair (A, S) of self mappings of a Menger space (X, F, t) is said to have the common limit range property with respect to the mappingS(denoted by (CLR_S))[12] if there exists a sequence {x_n} inXsuch that

   limn→∞Axn=limn→∞Sxn=z,  where  z∈S(X).

4. Two pairs (A, S) and (B, T) of self mappings of a Menger space (X, F, t) are said to have the common limit range property with respect to mappingsSandT [12] (denoted by (CLR_ST)) if there exists two sequences {x_n}, {y_n} inXsuch that

   limn→∞Axn=limn→∞Sxn=limn→∞Byn=limn→∞Tyn=z,  where  z∈S(X)∩T(X).

Definition 1.6

([15]). Self mappingsAandBof a Menger space (X, F, t) are said to be weakly compatible if they commute at their coincidence points, i.e., ifAx = Bxfor somex ∈ X, thenABx = BAx.

Lemma 1.7

LetA, B, S, T, LandMbe self mappings of a Menger space (X, F, t), with continuoust-norm witht(x, x) ≥ xfor allx ∈ [0, 1], satisfying the following conditions:

1. L(X) ⊆ ST(X) [resp. M(X) ⊆ AB(X)];

2. the pair (L, AB) satisfies the (CLR_AB) property[resp. the pair (M, ST) satisfies the (CLR_ST) property];

3. ST(X) is a closed subset ofX [resp. AB(X) is a closed subset ofX];

4. there exists an upper semicontinuous functionϕ : [0, ∞) → [0, ∞) withϕ(0) = 0 andϕ(x) < xfor allx > 0 such that

   FLp,Mq(ϕ(x))≥min{FABp,Lp(x),FSTq,Mq(x),FSTq,Lp(βx),FABp,Mq((1+β)x),FABp,STq(x)}(1)

   for allp, q ∈ X, β ≥ 1 andx > 0.

   Then the pairs (L, AB) and (M, ST) share the (CLR_(AB)(ST)) property.

Proof

Since the pair (L, AB) satisfies the (CLR_AB) property, there exists a sequence {x_n} in X such that

In view of (i) and (iii), for {x_n} ⊂ X, there exists a sequence {y_n} ⊂ X such that Lx_n = STy_n. It follows that

Therefore, it suffices to prove that limn→∞My_n = z. In fact, by (iv), putting p = x_n, q = y_n, we can obtain that

If min{F_{ABxn, Lxn}(x), F_{Lxn, Myn}(x)} = F_{Lxn, Myn}(x). Since ϕ(x) < x for all x > 0 and F is non-decreasing, then we get F_{Lxn, Myn}(ϕ(x)) < F_{Lxn, Myn}(x) which is a contradiction.

Therefore min{F_{ABxn, Lxn}(x), F_{Lxn, Myn}(x)} = F_{ABxn, Lxn}(x). It follows that

Letting n → ∞ in above inequality, then we have F_{ABxn, Lxn}(x) → F_{z, z}(x) = 1. Thus, limn→∞My_n = z. It yields that

i.e., the pairs (L, AB) and (M, ST) share the (CLR_(AB)(ST)) property.□

Remark 1.8

It can be pointed that Lemma 1.7 generalizes Lemma 3.2 in [12], from four self-mappings to six self-mappings. Simultaneously, it is straight forward to notice that Lemma 1.7 improves Lemma 1 of [13] from symmetric spaces to Menger spaces.

Lemma 2.1

([17]). Suppose that the functionϕ: [0, ∞) → [0, ∞) is upper semicontinuous withϕ(0) = 0 andϕ(x) < xfor allx > 0. Then there exists a strictly increasing continuous function α : [0, ∞) → [0, ∞) such thatα(0) = 0 andϕ(x) ≤ α(x) < xfor allx > 0. The functionαis invertible and for anyx > 0, limn→∞α^–n = ∞, whereα^–ndenotes then-th iterates ofα^–1andα^–1denotes the inverse ofα.

Lemma 2.2

([17]). Suppose that (X, F) is a PM-space andα: [0, ∞) → [0, ∞) is a strictly increasing function satisfyingα(0) = 0 andα(x) < xfor allx > 0. Ifx, yare two members inXsuch that

for allϵ > 0, thenx = y.

Theorem 2.3

LetA, B, S, T, LandMbe self mappings of a Menger space (X, F, t), with continuoust-norm witht(x, x) ≥ xfor allx ∈ [0, 1]. Suppose that the inequality(1)of Lemma 1.7 holds. If the pairs (L, AB) and (M, ST) share the (CLR_(AB)(ST)) property, then (L, AB) and (M, ST) have a coincidence point each.

Moreover, if

1. both the pairs (L, AB) and (M, ST) are weakly compatible.

2. AB = BA, LA = AL, MS = SMandST = TS.

   ThenA, B, S, T, LandMhave a unique common fixed point.

Proof

Since the pairs (L, AB) and (M, ST) share the (CLR_(AB)(ST)) property, there exist two sequences {x_n}, {y_n} in X such that

Since z ∈ ST(X), there exists a point u ∈ X such that STu = z. Putting p = x_n and q = u in inequality (1), it yields that

Letting n → ∞, we obtain that

From Lemma 2.1, there exists a strictly increasing continuous function α : [0, ∞) → [0, ∞) such that α(0) = 0 and ϕ(x) ≤ α(x) < x for all x > 0. Therefore, F_{z, Mu}(α(x)) ≥ F_{z, Mu}(ϕ(x)) ≥ F_{z, Mu}(x), for all x > 0. By Lemma 2.2, we obtain that z = Mu. Hence, z = Mu = STu, which shows u is a coincidence point of the pair (M, ST).

As z ∈ AB(X), there exists a point v ∈ X such that ABv = z, putting p = v, q = y_n in inequality (1), we have

Letting n → ∞, we obtain that

From Lemma 2.1, there exists a strictly increasing continuous function α : [0, ∞) → [0, ∞) such that α(0) = 0 and ϕ(x) ≤ α(x) < x for all x > 0. Therefore, F_{Lv, z}(α(x)) ≥ F_{Lv, z}(ϕ(x)) ≥ F_{Lv, z}(x), for all x > 0. By Lemma 2.2, we obtain that z = Lv. Hence, z = ABv = Lv, which shows v is a coincidence point of the pair (L, AB).

Since the pair (M, ST) is weakly compatible, and by the previous proof, z = Mu = STu, then MSTu = STMu, it yields that Mz = STz. And since the pair (L, AB) is weakly compatible, and by the previous proof, z = ABv = Lv, then LABv = ABLv, it yields that Lz = ABz. Letting p = z, q = u in inequality (1), we obtain:

From Lemma 2.1, there exists a strictly increasing continuous function α :[0, ∞) → [0, ∞) such that α(0) = 0 and ϕ(x) ≤ α(x) < x for all x > 0. Therefore, F_{Lz, Mu}(α(x)) ≥ F_{Lz, Mu}(ϕ(x)) ≥ F_{Lz, Mu}(x), for all x > 0. By Lemma 2.2, we obtain that Lz = Mu. Therefore, Lz = Mu = z. Thus, z = Lz = ABz.

Sequentially, letting p = z, q = z in inequality (1), we obtain:

From Lemma 2.1, there exists a strictly increasing continuous function α : [0, ∞) → [0, ∞) such that α(0) = 0 and ϕ(x) ≤ α(x) < x for all x > 0. Therefore, F_{Lz, Mz}(α(x)) ≥ F_{Lz, Mz}(ϕ(x)) ≥ F_{Lz, Mz}(x), for all x > 0. By Lemma 2.2, we obtain Lz = Mz = z = STz = ABz. Hence, AB, ST, L and M have a common fixed point z.

Letting p = z, q = Sz in inequality (1), we obtain:

From Lemma 2.1, there exists a strictly increasing continuous function α :[0, ∞) → [0, ∞) such that α(0) = 0 and ϕ(x) ≤ α(x) < x for all x > 0. Therefore, F_{z, Sz}(α(x)) ≥ F_{z, Sz}(ϕ(x)) ≥ F_{z, Sz}(x), for all x > 0. By Lemma 2.2, we obtain z = Sz. Thus, z = Sz = STz = TSz = Tz.

Letting p = Az, q = z in inequality (1), we obtain:

From Lemma 2.1, there exists a strictly increasing continuous function α :[0, ∞) → [0, ∞) such that α(0) = 0 and ϕ(x) ≤ α(x) < x for all x > 0. Therefore, F_{Az, z}(α(x)) ≥ F_{Az, z}(ϕ(x)) ≥ F_{Az, z}(x), for all x > 0. By Lemma 2.2, we obtain z = Az. Hence, z = Az = ABz = BAz = Bz. Thus, combing with the above proof, we have z = Az = Bz = Lz = Mz = Sz = Tz.

Then, A, B, S, T, L and M have a common fixed point z.

(Uniqueness). Assume that t is another common fixed point of A, B, S, T, L and M. It follows that t = At = Bt = Lt = Mt = St = Tt. Letting p = z, q = t in inequality (1), we obtain:

It yields that F_{z, t}(ϕ(x)) ≥ F_{z, t}(x). From Lemma 2.1, there exists a strictly increasing continuous function α :[0, ∞) → [0, ∞) such that α(0) = 0 and ϕ(x) ≤ α(x) < x for all x > 0. Therefore, F_{z, t}(α(x)) ≥ F_{z, t}(ϕ(x)) ≥ F_{z, t}(x), for all x > 0. By Lemma 2.2, we obtain z = t. Thus, A, B, S, T, L and M have a unique common fixed point z.□

Corollary 2.4

LetA, S, LandMbe self mappings of a Menger space (X, F, t), with continuoust-norm witht(x, x) ≥ xfor allx ∈ [0, 1]. Suppose that the inequality

holds. If the pairs (L, A) and (M, S) share the (CLR_(AS)) property, then (L, A) and (M, S) have a coincidence point each. Moreover, if both the pairs (L, A) and (M, S) are weakly compatible, LA = AL, MS = SM, thenA, S, LandMhave a unique common fixed point.

Remark 2.5

Theorem 2.3 generalizes Theorem 3.3 of [15]. Here, completeness of Menger space (X, F, t), the containment ofL(X) ⊆ ST(X), M(X) ⊆ AB(X) and the closure ofAB(X) orST(X) can be replaced by (CLR_(AB)(ST)) property of the pairs (L, AB) and (M, ST). Simultaneously, BL = LB, MS = SMcan be replaced withAL = LA, MS = SM. Of course, Theorem 2.3 also improves Theorem 3.4 of [15], the containment ofL(X) ⊆ ST(X), M(X) ⊆ AB(X) and the closure ofAB(X) and the property (E.A.) of (M, ST) or the closure ofST(X) and the property (E.A.) of (L, AB) can be removed, BL = LB, MS = SMcan be replaced withAL = LA, MS = SM. Meanwhile, Theorem 2.3 improves results of [13] from symmetric spaces to Menger spaces. In other respect, Theorem 2.3 improves Theorem 3.4 of [12], from four self mappings to six self-mappings in Menger spaces. To above all, we can deduce that the inequality (1) is different from that of [12].

Example 2.6

LetX = [0, 3), with the metricddefined byd(x, y) = | x – y | and defineF_{x, y}(u) = H( u – d(x, y)) for allx, y ∈ X, u > 0(refer to [15, Example 3.2]). It is obviously that the spaceXis not complete, since it is not a closed interval in real numbers ℝ. We definet(a, b) = min {a, b} for alla, b ∈ [0, 1]. LetA, B, S, T, LandMbe self mappings onXdefined as

AndL(x) = M(x) = 2. By a simple calculation, we can check the conditions in Theorem 2.3 hold true.

1. Consider two sequences{xn}={1+1n} and {yn}={2−1n}.ThenLx_n = 2, ABx_n = A(2) = 2, My_n = 2, STyn=S(13(yn+4))=16(2−1n)+53=2−16nwhich consequently it yields that

   limn→∞Lxn=limn→∞ABxn=limn→∞STyn=limn→∞Myn=2, where 2∈AB(X)∩ST(X).

   Therefore, the pairs (L, AB) and (M, ST) have the (CLR_(AB)(ST)) property. It is obvious thatST(X)=[53,136)is not closed inX.

2. Check the inequality(1). Letϕ : [0, ∞] → [0, ∞] defined byϕ(t) = kt, k ∈ (0, 1) be an upper semicontinuous function withϕ(0) = 0 andϕ(t) < tfor allt > 0. For anyp, q ∈ ℝ andx > 0, we haveF_{Lp, Mq}(kx) = 1 and

   min{FABp,Lp(x),FSTq,Mq(x),FSTq,Lp(βx),FABp,Mq((1+β)x),FABp,STq(x)}=min{1,FSTq,Mq(x),FSTq,Lp(βx),1,FABp,STq(x)}=min{FSTq,Mq(x),FSTq,Lp(βx),FABp,STq(x)}=min{FSTq,2(x),FSTq,2(βx),F2,STq(x)}=FSTq,2(x)≤1.

   ThenF_{Lp, Mq}(kx) ≥ min{F_{ABp, Lp}(x), F_{STq, Mq}(x), F_{STq, Lp}(βx), F_{ABp, Mq}((1+β)x), F_{ABp, STq}(x)} holds forx, y ∈ X, β ≥ 1 andx > 0.

3. It is obviously thatL(x) = AB(x) = {2} for 1 ≤ x < 3, andL(AB)(x) = (AB)L(x) = {2}. Then the weakly compatibility of the pair (L, AB) is satisfied. AndM(x) = ST(x) = {2} forx = 1, andM(ST)(x) = 2 = ST(2) = (ST)M(x) forx = 1. Then the weakly compatibility of the pair (M, ST) is also satisfied.

4. AB = BA = {2}, ST = TS = 16x+53,LA = LA = {2}, andSM = MS = {2}.

Thus, all the conditions of Theorem 2.3 are satisfied, but 2 is a unique common fixed point ofA, B, S, T, LandM.

Theorem 2.7

LetA, B, S, T, LandMbe self mappings of a Menger space (X, F, t), with continuoust-norm witht(x, x) ≥ xfor allx ∈ [0, 1]. Suppose that the conditions(i)-(iv) of Lemma 1.7 hold. Then (L, AB) and (M, ST) have a coincidence point each.

Moreover, if

1. both the pairs (L, AB) and (M, ST) are weakly compatible.

2. AB = BA, LA = AL, MS = SMandST = TS.

ThenA, B, S, T, LandMhave a unique common fixed point.

Proof

Since the conditions (i)-(iv) of Lemma 1.7 hold, thus the pairs (L, AB) and (M, ST) have the (CLR_(AB)(ST)) property. The rest of proof can be completed along the routine of the proof of Theorem 2.3. In order to avoid tedious presentation, we omit the rest of proof.□

Example 2.8

Assume the same conditions of Example 2.6, except that

AndL(x) = M(x) = 2. First, we can check the conditions in Lemma 1.7.

1. L(X)=2,ST(X)=[43,229]. Thus,L(X)⊆ST(X).

2. Takex_n = 1 – 1/n ∈ X. Thenlimn→∞AB(x_n) = limn→∞AB(1 – 1/n) = {2} andlimn→∞L(x_n) = limn→∞L(1 – 1/n) = {2}. Therefore, limn→∞AB(x_n) = limn→∞L(x_n). It yields that the pair (L, AB) satisfies the property (E, A).

3. ST(X) = [43,229].It is a closed interval in ℝ, of course, it is closed subset ofX.

4. Check the inequality(1). Letϕ: [0, ∞] → [0, ∞] defined byϕ(t) = kt, k ∈ (0, 1) be an upper semicontinuous function withϕ(0) = 0 andϕ(t) < tfor allt > 0. For anyp, q ∈ ℝ andx > 0, we haveF_{Lp, Mq}(kx) = 1 and

   min{FABp,Lp(x),FSTq,Mq(x),FSTq,Lp(βx),FABp,Mq((1+β)x),FABp,STq(x)}=min{1,FSTq,Mq(x),FSTq,Lp(βx),1,FABp,STq(x)}=min{FSTq,Mq(x),FSTq,Lp(βx),FABp,STq(x)}=min{FSTq,1(x),FSTq,1(βx),F1,STq(x)}=FSTq,1(x)≤1.

ThenF_{Lp, Mq}(kx) ≥ min{F_{ABp, Lp}(x), F_{STq, Mq}(x), F_{STq, Lp}(βx), F_{ABp, Mq}((1 + β)x), F_{ABp, STq}(x)} holds forx, y ∈ X, β ≥ 1 andx > 0.

Besides, we should check weak compatibility of (M, ST). M(x) = ST(x) = {2} forx = 2, andM(ST)(x) = M(2) = 2 = ST(2) = (ST)M(x) for x = 2. Then the weakly compatibility of the pair (M, ST) is also satisfied.

At the last, ST=TS=43,x=0,229,x∈(0,1],4x+109,x∈(1,3).and SM=MS={43}.

Thus all the conditions of Theorem 2.7 are satisfied. From Theorem 2.7, A, B, S, T, LandMhave a unique common fixed point inX. In fact, by the definition ofA, B, S, T, LandM, 2 is the unique common fixed point ofA, B, S, T, LandMinX.

Theorem 2.9

LetA, B, S, T, LandMbe self mappings of a Menger space (X, F, t), with continuoust-norm witht(x, x) ≥ xfor allx ∈ [0, 1]. Suppose that the inequality(1)of Lemma 1.7 and the following hypotheses hold:

1. the pairs (L, AB) and (M, ST) have the common property (E.A.);

2. ST(X) andAB(X) is closed subset ofX.

Then (L, AB) and (M, ST) have a coincidence point each.

Moreover, if

1. both the pairs (L, AB) and (M, ST) are weakly compatible.

2. AB = BA, LA = AL, MS = SMandST = TS.

ThenA, B, S, T, LandMhave a unique common fixed point.

Proof

If the pairs (L, AB) and (M, ST) have the common property (E.A.), then there exist two sequences {x_n} and {y_n} in X such that

Since ST(X) is closed, then limn→∞STy_n = z = STu for some u ∈ X. And AB(X) is closed, then limn→∞ABx_n = z = ABv for some v ∈ X. The rest of the proof can runs on the lines of Theorem 2.3.□

Corollary 2.10

The result of Theorem 2.9 holds if condition (b’) is substituted for condition (b):

(b’)L(X) ⊆ ST(X) andM(X) ⊆ AB(X) where⋅denoted the closure.

Corollary 2.11

The result of Theorem 2.9 holds if condition (b”) is substituted for condition (b):

(b”)L(X) andM(X) is closed subset ofX, andL(X)⊆ ST(X) andM(X) ⊆ AB(X).

Example 2.12

Assume the same conditions ofExample 2.6hold, except that

ST(X)={83}∪{43}∪(25,4)is not closed subset ofX, but conditions (b’) and (b”) of Corollary 2.10 and Corollary 2.11 are satisfied, 2 is a unique common fixed point ofA, B, S, T, LandM.

Remark 2.13

Theorem 2.9 improves Theorem 3.4 in [15]. Here, containment ofL(X) ⊆ ST(X), M(X) ⊆ AB(X), and the closure ofST(X), property (E.A) of (L, AB) are replaced by the closure ofST(X) andAB(X), and the common property (E.A) of two pairs (L, AB) and (M, ST). Of course, LB = BL, MT = TMare also replaced byLA = AL, MS = SM. Indeed, the common property (E.A) of two pairs (L, AB) and (M, ST) can be deduced from containment ofL(X) ⊆ ST(X) and property (E.A) of (L, AB).

Theorem 2.14

LetA, B, S, T, LandMbe self mappings of a Menger space (X, F, t), with continuoust-norm witht(x, x) ≥ xfor allx ∈ [0, 1]. Suppose that the inequality(1)of Lemma 1.7 and the following hypotheses hold:

1. L(X) ⊆ ST(X);

2. ST(X) is closed inXand (L, AB) satisfies the property (E.A).

Then (L, AB) and (M, ST) share the common property (E.A).

Proof

Since (L, AB) satisfies the property (E.A), there exists a sequence {x_n} ⊂ X such that

Since L(X) ⊆ ST(X), for each x_n, there exists a corresponding y_n ∈ X such that Lx_n = STy_n. Therefore, we have

It suffices to show that limn→∞My_n = z. Substituting p = x_n, q = y_n in inequality (1), we obtain

Letting n → ∞, we obtain that F_{ABxn, Lxn}(x) → F_{z, z}(x) = 1. So,

Thus, (L, AB) and (M, ST) share the common property (E.A).□

Remark 2.15

Theorem 2.14 shows that our common property (E.A) of two pairs (L, AB) and (M, ST) is weaker than containment ofL(X) ⊆ ST(X) and property (E.A) of (L, AB). It is namely thatTheorem 2.9is indeed a generalization of Theorem 3.4 in [15].

Theorem 2.16

Let{Ai}i=1m,{Br}r=1n,{Sk}k=1e,{Th}h=1f,{Lj}j=1c and {Mv}v=1dbe six finite families of self mappings of a Menger space (X, F, t), with continuoust-norm witht(x, x) ≥ xfor allx ∈ [0, 1] whereA = A₁A₂ ⋯ A_m, B = B₁B₂ ⋯ B_n, S = S₁S₂ ⋯ S_e, T = T₁T₂ ⋯ T_f, L = L₁L₂ ⋯ L_c and M = M₁M₂ ⋯ M_d. Suppose that the inequality(1)of Lemma 1.7 holds. If the pairs (L, AB) and (M, ST) share the (CLR_(AB)(ST)) property, then (L, AB) and (M, ST) have a coincidence point each.

Moreover, if

1. both the pairs (L, AB) and (M, ST) are weakly compatible.

2. AB = BA, LA = AL, MS = SMandST = TS.

ThenA, B, S, T, LandMhave a unique common fixed point.

Proof

The proof can be completed on the lines of Theorem 4.2 in Imdad et al. [12].□

Corollary 2.17

LetA, B, S, T, LandMof self mappings of a Menger space (X, F, t), with continuoust-norm witht(x, x) ≥ xfor allx ∈ [0, 1]. Suppose that

1. the pairs (L^c, A^mBⁿ) and (M^d, S^eT^f) share the (CLR_{(A^mBⁿ)(S^eT^f)}) property,

2. there exists an upper semicontinuous functionϕ : [0, ∞) → [0, ∞) withϕ(0) = 0 andϕ(x) < xfor allx > 0 such that

   FLcp,Mdq(ϕ(x))≥min{FAmBnp,Lcp(x),FSeTfq,Mdq(x),FSeTfq,Lcp(βx),FAmBnp,Mdq((1+β)x),FAmBnp,SeTfq(x)},

   for allp, q ∈ X, β≥ 1 andx > 0. Then (L^c, A^mBⁿ) and (M^d, S^pT^q) have a coincidence point each.